In triangle ABC we have angle A=100 and angles B=C=40. The side AB is produced to a point D so that B lies between A and D and AD=AC. Find angle BCD

Dear student,

Since AD=ACACD=CDA   --[angles opposite to equal sides are equal]In ACDACD+CDA+CAD=180°x+x+100°=180°2x=180°-100°2x=80°x=80°2x=40°ACD=40°But this is not possible as ACB=40°   (given)Recheck your post and get back to us for meanigful help.
Regards

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